GIVEN :
a = 1/√10 ( 3i + k) and
b = 1/7 ( 2i + 3j - 6k)
TO FIND :
( 2a- b) . [ ( a x b ) x ( a + 2b)]
SOLUTION :
◆Going with the equation given,
( 2a - b) . [ ( a x b ) x ( a + 2b)]
= (2a - b) [( a×b×a) + 2(a×b)×b]
◆BAC - CAB RULE,
A×B×C = B( A.B ) - C(A.B )
= (2a- b ) [ (b (a.a ) - a (a.b ) + 2b ( a.b) -2b (a.b]
Solving further
= (2a - b )(b - 2a)
= -4a.a -b.b
=-5.
Answer:
( 2 - b) . [ ( a x b ) x ( a + 2b)] = -5
Hoped I helped
Bonjour,
answer ≈ 522.92
working
A= 2πrh+2πr^2
=2 x π x 4 x 18+2 x π x 4^2
≈552.92031
Answer:
5/10
Step-by-step explanation:
5+5=10. so 5 empty 0/10 5 shaded 5/10
Answer:
A
i had this question it's right
In a quadratic equation with the general formula of:
ax^2 + bx + c = 0
The discriminant is equal to b^2 - 4(a)(c). If the answer is a perfect square, then there are two real numbers. If not, then there are no real number root.
The discriminant for this equation is
(-6)^2 - 4(3)(1) = 24
Since 24 is not a perfect square, there are no real number roots.